Chapter U4.13, Problem 5E

Interpretation Introduction

Interpretation: The given salt solutions should be listed in increasing molarity.

Concept Introduction: The molar concentration of a solute in a solution can be expressed in terms of molarity as moles of solute present in per liter of solution. Mathematically,

  $\text{Molarity = }\frac{\text{number of moles of solute}}{\text{Volume of solution (L)}}$ -(1)

Answer to Problem 5E

The list of solution in increasing order of molarity is:

  $\text{1}\text{.0 mol NaCl per 10 L < }4.0\text{ mol NaCl per 8}\text{.0 L < 6}\text{.0 mol NaCl per 6}\text{.0 L}$ .

Explanation of Solution

Given:

  $4.0\text{ mol NaCl per 8}\text{.0 L}$ , $\text{6}\text{.0 mol NaCl per 6}\text{.0 L}$ and $\text{1}\text{.0 mol NaCl per 10 L}$ .

Substituting the given values in equation (1) to determine the molarity of the solution as:

  $\text{Molarity = }\frac{\text{number of moles of solute}}{\text{Volume of solution (L)}}$

• For $4.0\text{ mol NaCl per 8}\text{.0 L}$:

  $\begin{array}{l}\text{Molarity = }\frac{\text{4}\text{.0 mol}}{\text{8}\text{.0 L}}\\ \text{Molarity = }0.5\text{ mol/L}\end{array}$

• For $\text{6}\text{.0 mol NaCl per 6}\text{.0 L}$:

  $\begin{array}{l}\text{Molarity = }\frac{\text{6}\text{.0 mol}}{\text{6}\text{.0 L}}\\ \text{Molarity = 1}\text{.0 mol/L}\end{array}$

• For $\text{1}\text{.0 mol NaCl per 10 L}$:

  $\begin{array}{l}\text{Molarity = }\frac{\text{1}\text{.0 mol}}{\text{10}\text{.0 L}}\\ \text{Molarity = 0}\text{.1 mol/L}\end{array}$

The molarity of all the three solutions is:

The molarity of $4.0\text{ mol NaCl per 8}\text{.0 L}$ is $\text{0}\text{.5 M}$ .

The molarity of $\text{6}\text{.0 mol NaCl per 6}\text{.0 L}$ is $\text{1}\text{.0 M}$ .

The molarity of $\text{1}\text{.0 mol NaCl per 10 L}$ is $\text{0}\text{.1 M}$ .

Hence, the list of solution in increasing order of molarity is:

  $\text{1}\text{.0 mol NaCl per 10 L < }4.0\text{ mol NaCl per 8}\text{.0 L < 6}\text{.0 mol NaCl per 6}\text{.0 L}$ .